Lable point B at (2,-3/2)

galben [10]

These are two questions and two answers.

Question 1) Which of the following polar equations is equivalent to the parametric equations below?

 x=t²
y=2t

Answer: option A.) r = 4cot(theta)csc(theta)


Explanation:

1) Polar coordinates ⇒ x = r cosθ and y = r sinθ

2) replace x and y in the parametric equations:

r cosθ = t²
r sinθ = 2t

3) work r sinθ = 2t

r sinθ/2 = t
(r sinθ / 2)² = t²

4) equal both expressions for t²

r cos θ = (r sin θ / 2 )²

5) simplify

r cos θ = r² (sin θ)² / 4

4 = r (sinθ)² / cos θ

r = 4 cosθ / (sinθ)²

r = 4 cot θ csc θ ↔ which is the option A.

Question 2) Which polar equation is equivalent to the parametric equations below?

 x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)

Answer: option B) r = sinθ + 1

Explanation:

1) Polar coordinates ⇒ x = r cosθ, and y = r sinθ

2) replace x and y in the parametric equations:

a) r cosθ = sin(θ)cos(θ)+cos(θ)
 b) r sinθ =sin²(θ)+sin(θ)

3) work both equations

a) r cosθ = sin(θ)cos(θ)+cos(θ) ⇒ r cosθ = cosθ [ sin θ + 1] ⇒ r = sinθ + 1

 b) r sinθ =sin²(θ)+sin(θ) ⇒ r sinθ = sinθ [sinθ + 1] ⇒ r = sinθ + 1


Therefore, the answer is r = sinθ + 1 which is the option B.

6 0

2 years ago

Read 2 more answers

Find the gain percent in the following: Cost Price Rs 280, (a) Selling Price Rs 290

Nikolay [14]

Answer:

3.57

Step-by-step explanation:

P=SP -CP

=290-280=10

P%= $\frac{P*100}{CP}$

= $\frac{10*100}{280}$

= $\frac{1000}{280}$

=3.57

4 0

3 years ago

HELP ME PLS GRADES ARE DUE TODAY

damaskus [11]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

Most of this exercise is looking at different ways to identify the slope of the line. The first attachment shows the corresponding "run" (horizontal change) and "rise" (vertical change) between the marked points.

In your diagram, these values (run=1, rise=-3) are filled in 3 places. At the top, the changes are described in words. On the left, they are described as "rise" and "run" with numbers. At the bottom left, these same numbers are described by ∆y and ∆x.

The calculation at the right shows the differences between y (numerator) and x (denominator) coordinates. This is how you compute the slope from the coordinates of two points.

If you draw a line through the two points, you find it intersects the y-axis at y=4. This is the y-intercept that gets filled in at the bottom. (The y-intercept here is 1 left and 3 up from the point (1, 1).)